How do you find the indefinite integral of $int (x+1/x)^2dx$ ?

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Answer 1 · 2017-01-17T23:49:06

$int(x+1/x)^2dx=x^3/3+2x-1/x+C$

Note that:

$(x+1/x)^2=(x+1/x)(x+1/x)$

$color(white)((x+1/x)^2)=x^2+2x(1/x)+(1/x)^2$

$color(white)((x+1/x)^2)=x^2+2+x^-2$

So:

$I=int(x+1/x)^2dx=intx^2dx+2intdx+intx^-2dx$

Using $intdx=x$ and $intx^ndx=x^(n+1)/(n+1)$ we obtain:

$I=x^3/3+2x+x^-1/(-1)+C$

$I=x^3/3+2x-1/x+C$

How do you find the indefinite integral of int (x+1/x)^2dxint (x+1/x)^2dx?